三角函数题 求解 要过程

sin2t+sint=0 求t 范围(0,2π）
sint(2cost+1)=0
sint=0,t在范围(0,2π）无解
2cost+1=0,
t=2π/3或t=4π/3

2tanx-sec^2(x)=0 求x
(2sinxcosx-1)/cos^2x=0,
sin2x=1,
x=π/4,或x=5π/4

2sin^2(u)=1-sinu 求u
2sin^2(u)+sinu-1=0
(2sinu-1)(sinu+1)=0
2sinu+1=0或sinu+1=0
sinu=-1/2,x=7π/6或x=11π/6
sinu=-1,x=3π/2